Question

If ${\sin }^2{5}^0+{\sin }^2{10}^0+{\sin }^2{15}^0+.....+{\sin }^2{90}^0=$

• A. 9
• B. 2
• C. 3
• D. 10

Answer 1

The correct option is A 9

It $si{n}^2{5}^0+si{n}^2{10}^0+si{n}^2{15}^0+...si{n}^2{90}^0$

Given $si{n}^2{5}^0+si{n}^2{10}^0+si{n}^2{5}^0+si{n}^2{20}^0+si{n}^2{25}^0+si{n}^{2}30+si{n}^2{35}^0+$

$si{n}^2{40}^0+si{n}^2{45}^0+si{n}^2{50}^2+si{n}^2{55}^0+si{n}^2{60}^0+si{n}^2{65}^0+si{n}^2{70}^0+si{n}^2{75}^0$

$si{n}^2{80}^0+si{n}^2{85}^0+si{n}^2{90}^0$

we know that

$si{n}^2(90-\theta )=co{s}^2\theta$

then $si{n}^2\theta +co{s}^2\theta =1$

Let us arrange the given question in the form of cos

$si{n}^2(90-85)+si{n}^2(90-80)+si{n}^2(90-75)+...+si{n}^2{85}^0+sin{90}^0$

$\Rightarrow co{s}^{2}85+co{s}^{2}80+co{s}^{2}75+co{s}^{2}70+co{s}^{2}65+co{s}^2{60}^0+$

$co{s}^2{55}^0+co{s}^{2}50+co{s}^2{45}^0+si{n}^2{50}^0+si{n}^2{55}^0+...+si{n}^{2}85+co{s}^{2}45$

$\Rightarrow (co{s}^{2}85+si{n}^{2}85)+(co{s}^{2}80+si{n}^{2}80)+(co{s}^{2}75+si{n}^{2}75)+$

$(co{s}^{2}70+si{n}^{2}70)+(co{s}^{2}85+si{n}^{2}65)+(co{s}^{2}60+si{n}^{2}60)+$

$(co{s}^{2}55+si{n}^{2}55)+(co{s}^{2}80+si{n}^2{80}^0)+(co{s}^{2}45+si{n}^{2}45)$

$\Rightarrow 1+1+1+1+1+1+1+1+1=9$

$\therefore si{n}^2{8}^0+si{n}^2{10}^0+si{n}^2{15}^0+...+si{n}^2{90}^0=9$